Question

If the position vectors of the point $A,B$ and $C$ are $-2i+j-k,-4i+2j+2k$ and $6i-3j-13k$ respectively and $AB=\lambda AC$, then find the value of $\lambda$.

Answer 1

We have,

$A=-2i+j-k$

$B=-4i+2j+2k$

$C=6i-3j-13k$

Now,

$AB=PositionvectorofB-PositionvectorofA$

$=(-4i+2j+2k)-(-2i+j-k)$

$=-2i+j+3k$

Similarly,

$AC=PositionvectorofC-PositionvectorofA$

$=(6i-3j-13k)-(-2i+j-k)$

$=8i-4j-12k$

Since,

$AB=\lambda AC$

$-2i+j+3k=\lambda (8i-4j-12k)$

$\lambda =-\frac{1}{4}$

Hence, this is the answer.